The Unapologetic Mathematician

Mathematics for the interested outsider

Mean Convergence Properties of Integrals

Okay, we’ve got our general definition of integrable functions, and we’ve reestablished a bunch of our basic properties in this setting. Let’s consider some properties that involve the L^1 norm.

First off, the basic definitions of the L^1 norm carries across: we set

\displaystyle\lVert f\rVert_1=\int f\,d\mu

and we say that a sequence \{f_n\} of integrable functions is Cauchy in the mean or is mean Cauchy if \lVert f_n-f_m\rVert_1 goes to zero for sufficiently large m and n. We immediately find that any sequence that is Cauchy in the mean is also Cauchy in measure, since our earlier proof didn’t really depend on the functions being simple.

We also have a couple more results whose proofs don’t really depend on the simplicity of f_n, and so these carry across without change. If \{f_n\} is mean Cauchy, with indefinite integrals \{\nu_n\}, then the collection of set functions \{\nu_n\} is uniformly absolutely continuous. Further, we can define the limiting set function


for every measurable E\subseteq X, and we find that this set function \nu is finite-valued and countably additive.

So now we can formally define the obvious notion: the sequence \{f_n\} of integrable functions “converges in the mean” or is “mean convergent” to an integrable function f if \lVert f_n-f\rVert_1 goes to zero as n goes to \infty. And, as we might expect, if \{f_n\} converges in the mean to f then it converges in measure as well.

Indeed, for every \epsilon>0 we define

\displaystyle E_n=\left\{x\in X\big\vert\lvert f_n-f\rvert\geq\epsilon\right\}

and then we see that

\displaystyle\int\lvert f_n-f\rvert\,d\mu\geq\int\limits_{E_n}\lvert f_n-f\rvert\,d\mu\geq\epsilon\mu(E_n)

Thus \mu(E_n) must go to 0 as n\to\infty, and so \{f_n\} converges in measure to f.

June 4, 2010 - Posted by | Analysis, Measure Theory


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